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operations C and C, are theoretically different, unless it be that in performing C, there is one degree more of freedom than in performing C1. But this is so also in the case of the two operations denoted by 2R,; for the ruler can be placed in coincidence with one point by a motion either of translation or of rotation or of both, while it can be placed in coincidence with the other point, the coincidence with the first being maintained, only by rotation.

In the case of the two operations denoted by 2C, it is clear also that there is less freedom in placing the second point of the compasses than there is in placing the first, and hence if, for the sake of convenience, two operations which are not precisely identical may be denoted by the reduplication of the same symbol, there does not seem to be any imperative reason why the operations C, and C should not be regarded as equivalent. The fact also that in estimating the simplicity and exactitude of constructions the symbol C2 rarely occurs, and the manifest advantage of having only four units instead of five have induced me to propose the following modification of Mr Lemoine's scheme :

To place the edge of the ruler in coincidence with
one point

R1

To place the edge of the ruler in coincidence with
two points.....

2R1

To draw a straight line

R2

To put one point of the compasses on a determinate
point

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To put the points of the compasses on two deter

minate points

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To describe a circle

C2

On another matter (of small importance) I have ventured to differ from Mr Lemoine.

Given A, B, C, the three vertices of a triangle, to construct the triangle.

Mr Lemoine estimates this operation as 6R1+3R. I have estimated it as 4R1+3R2. To put the ruler in contact with A, B is 2R,; to draw AB is R. Now as the ruler is in contact with B, I estimate the putting of it in contact with B and C as

only an additional R1; to draw BC is R2. Again as the ruler is in contact with C, I estimate the putting of it in contact with C and A as another R,; to draw CA is R-in all, 4R, +3R2.

The following remarks are extracted from one of Mr Lemoine's letters:

Geometrography may be divided into several branches.

(1) That of the canonical geometry of the straight line and the circle, the only instruments being the ruler and the compasses.

(2) Add the carpenter's square, with two new symbols. This branch may be applied especially to descriptive geometry. (3) Add graduated rulers, for application to graphical statics. (4) The geometrography of the ruler alone.

(5) The geometrography of the compasses alone.*

A sub-section may be made of the geometrography of the ruler and one single opening of the compasses. †

In what follows, the notation I. 1, etc., denotes Euclid's Elements, Book First, Proposition First, etc. It will be seen that, except in the fourth book, Euclid does not group his problems together.

I. 1.

To describe an equilateral triangle on a given finite straight line. 3R1+2R2+3C1 + 2C2

1

Simplicity 10; exactitude 6; lines 2; circles 2.

I. 2.

From a given point to draw a straight line equal to a given straight line.

5R1+3R2+7C1+4C2

The problem may be solved with much less complication, namely,

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From the greater of two given straight lines to cut off a part equal to the less.

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*See Mascheroni's Geometria del compasso (1795).

+ See Proceedings of the Edinburgh Mathematical Society, V. 2-22 (1887).

The problem may be solved with much less complication, namely,

3C1 + C2

I. 9.

To bisect a given rectilineal angle.

2R1 + R2+4C1 + 3C1⁄2

In this estimate the operations for drawing the sides of the equilateral triangle which occurs in Euclid's construction are omitted. The construction may be effected by

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Some of Euclid's operations are not counted, as they are needed only for the demonstration. The construction may be effected by

2R1 + R2+ 2C1 + 2C2

I. 11.

To draw a straight line perpendicular to a given straight line from a given point in the same.

2R1 + R2+4C1 + 3C2

The construction may be effected by

2R1 + R2+ 3C1 + 3C2

Or thus:

FIGURE 1.

Let AB be the straight line, C the point in it.

Take any point D outside AB; with D as centre and DC as radius describe a circle cutting AB again at E.

Join ED, and produce it to meet the circle at F; join FC.

3R, +2R2+ C1+C2

I. 12.

To draw a straight line perpendicular to a given straight line from a given point outside it.

2R1 + R2+5C1 + 3C2

From the way in which Euclid describes his construction, the formula for it would be

4R1+2R2+5C1 + 3C2

But if the construction be fully carried out it will be seen that the drawing of the final straight line is unnecessary. Hence the formula is as first stated.

The construction may be effected by

2R1 + R2 + 3C1 + 3C2

Or thus:

FIGURE 2.

Let AB be the straight line, C the point outside it.

Take any point D in AB; with D as centre and DC as radius describe a circle cutting AB at E.

With E as centre and EC as radius describe a circle cutting the previous one again at F; join FC.

2R1+R+4C1+2C2

I. 22.

To make a triangle the sides of which shall be equal to three given straight lines.

Euclid does not use any of the given straight lines as a side of the triangle.

3R1+3R2+9C1 + 4C2

I. 23.

At a given point in a given straight line to make an angle equal to a given angle.

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Through a given point to draw a straight line parallel to a given

straight line.

3R1+2R2+9C1+3C2

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The following method is due to Mr Gaston Tarry.

FIGURE 3.

Let A be the given point, BC the given straight line.

Draw any circle passing through A and cutting BC at D and E. With E as centre and radius AD describe a circle to cut the previous one at F. Join AF.

2R+R+4C, +202

I. 42.

To describe a parallelogram that shall be equal to a given triangle and have one of its angles equal to a given angle.

Euclid constructs his parallelogram on the half of one of the sides of the triangle.

10R1+6R2+30C1 +11C2

The construction may be effected by

8R1+4R2+15C1 + 9C2

I. 44.

To a given straight line to apply a parallelogram which shall be equal to a given triangle and have one of its angles equal to a given angle.

28R1+17R+81C1+28C2

I. 45.

To describe a parallelogram equal to a given rectilineal figure and having an angle equal to a given angle.

Euclid takes a quadrilateral for the given rectilineal figure.

40R2+24R2+1110, +39C2

I. 46.

To describe a square on a given straight line.

10R1+6R2+2401 + 10C,

The construction may be effected by

6R1 +3R2+7C1 +5C2

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